Shear ultimate limit state
EUROCODE-2

Shear ultimate limit state

DATA
Dimensions (?)
b(cm)     h(cm)     c(mm)
  • b: width section. Valid values from 10 to 150
  • h: depth section. Valid values from 10 to 150
  • c: Concrete cover. Valid values from 10 to h/3
Concrete strength (?)
fck (MPa)       γc       αcc
  • fck: the characteristic compressive cylinder strength of concrete. Valid values from 20 to 90
  • γc: Partial factor for concrete for ultimate limit state. National choice 2.4.2.4(1). Valid values from 1 to 2
  • αcc: National choice 3.1.6 (1)P: Coefficient of fatigue strength of the concrete. Recommended value is 1.0. Valid values from 0.5 to 1
Reinforcement strength
fyk(MPa)   (?)   γs
  • fyk: the characteristic strength of reinforcement. Valid values from 400 to 600
  • γs: Partial factor for reinforcement for ultimate limit state. National choice 2.4.2.4(1). Valid values from 1 to 1.8

shear reinforcement

REINFORCEMENT
Member with shear reinforcement
Transverse reinforc. (?)
Asw(cm2)     s(cm)     α(osex)
  • Asw: Cross-sectional area of shear reinforcement. Valid values from Ast,min >0 to Asw,max (Φ32 bars in width section separated 5cm).
  • s: The spacing of the stirrups. Valid values from 5 to 2h.
  • α: Angle between the reinforcements and the member′s axis. Valid values from 45 to 90.

You can enter the area directly or through the following data:

typeNum Φ
1
2
Tensile reinforcement
As1 (cm2 (?)  Φmax
  • As1: Cross-sectional area of tensile reinforcement. Valid values from 0 to As1,max (layer Φ40 with bar spacing s=5 cm).
  • Φmax: max. nominal diameter of a reinforcing bar. Valid values from 5 to 40

You can enter the area directly or through the following data:

typeNum Φ
1
2
Compression reinforc.
As2 (cm2 (?)
  • As2: Cross-sectional area of compression reinforcement. Valid values from 0 to As2,max (0.5·fcd·Ac/fyd).

You can enter the area directly or through the following data:

typeNum Φ
1
2

INTERNAL FORCES
Shear force VEd(Kn) (?)

Design shear force in the section considered. Valid values from 0 to 0.5·fcd·Ac

Axial force NEd(Kn) (?)

Design axial force in the section considered. Valid values from 0 to fcd·Ac

Inner lever arm (?)
z = · d

Inner lever arm corresponding to the bending moment. For concrete without axial force, use z = 0,9·d. Valid values from 0,9·d to 0,6·d


VALUES FOR USE IN A COUNTRY
Use values recommended
Value vmin (?)
·k3/2·fck1/2

Value design shear (see clause 6.2.2(1)
The recommended value is 0.035·k3/2·fck1/2(Valid values from (0.01 to 0.09)·k3/2·fck1/2)

CRd,c     (?)     k1
/γc    

Values design shear (see clause 6.2.2(1)

  • CRd,c: The recommended value is 0.18/γc (Valid values from (0.10 to 0.25)/γc)
  • k1: The recommended value is 0.15 (Valid values from 0.1 to 0.2).

min(cotθ)   (?)   max(cotθ)
≤ cotθ ≤

θ is the angle between the concrete compression strut and the member axis.

  • min(cotθ): The recommended value is 1. Valid values from 0.2 to 1
  • max(cotθ): The recommended value is 2.5. Valid values from 1 to 4
Value v1 (?)
·0.6·(1-fck/250)

Value design shear (see clause 6.2.3(3)
The recommended value is 0.6·(1-fck/250). You can multiply this value by a factor between 0.1 and 3

Value acw (?)
·Recom_value

Value design shear (see clause 6.2.3(3). The recommended value is

  • 1 for non-prestressed structures
  • (1+σcp/fcd), for 0<σcp≤0.25 fcd
  • 1.25, for 0.25·fcdcp≤0.5·fcd
  • 2.5(1-σcp/fcd), for 0.5·fcdcp≤fcd

You can multiply this value by a factor between 0.1 and 3

 

RESULT

Design shear force Design shear strength Requirement
VEd (KN) VRd,max (KN) VRd (KN) VEd < VRd and VRd,max
200 305.01 239.77 OK



DETAILS OF CALCULATION

Notation and methodology according to clause 6.2 of EC2

VRd,max (Resistance crushing of compression struts)= 305.01 kN
VRd,max = αcw · bw · z · v1 · fcd·(cotθ + cotα) / (1+cot2θ)
VRd,max = 1·300·328·0.54·16.67·(2.5+0) / (1+2.52) = 305007 N
where:

  • fcd = αcc · fck / γc = 1 · 25 / 1.5 = 16.67 MPa
  • αcw = 1 · 1 = 1
    Case: non-prestressed structure or non-axial force
  • z = 0.9 · d = 32.76 cm
    d (effective depth) = h – c - Φmax/2 = 40 – 3 – 1.2/2 = 36.4 cm
  • v1 = 1·0.6·(1-25/250) = 0.54
  • cotθ = 2.5 (1 ≤ cotθ ≤ 2.5)
    (it get the maximum value of VRd,s that satisfies VEd ≤ VRd,max)

VRd (Tensile strength in the web) = 239.77 KN
VRd = max (VRd,c ; VRd,s), in which:

  • VRd,c (Resistance without shear reinforcement)= 53.56 kN
    VRd,c = [CRd,ck(100·ρ1·fck)1/3 + k1·σcp] bw·d
    VRd,c = [0.12·1.741·(100·0.0052·25)1/3 + 0.15·0]·300·364 = 53562 N
    with a minimum of
    VRd,c (min) = (vmin + k1·σcp) bw·d
    VRd,c (min) = (0.402 + 0.15·0)·300·364 = 43909 N
    where:

    • CRd,c = 0.18/1.5 = 0.12
    • k = min [1+(200/d)1/2 , 2] = 1.741
      1+(200/d)1/2 = 1+(200/364)1/2 = 1.741
    • ρ1 = min (As1/(b∙d) , 0.02) = 0.0052
      As1/(b∙d) = 5.65/(100∙36.4) = 0.0052
    • σcp = min [NEd/(b·h) ; 0.2·fcd] = min (0 ; 3.33) = 0 MPa
    • vmin = 0.035·k3/2·fck1/2 = 0.035·1.7413/2·251/2 = 0.402 MPa

  • VRd,s (Resistance with shear reinforcement)= 239.77 kN
    VRd,s = Asw/s · z · fywd · (cotθ+cotα)·sinα
    VRd,s = 101/15 · 32.76 · 434.78 · (2.5+0)·1 = 239765 N
    with fywd = fyk / γa = 500 /1.15 = 434.78 N/mm2